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Where in the question does it say the height cant be one of the sides. Any answer other than the one PHILIP gave is impossible.
I am afraid this question is an old chestnut which is intended to test logic as well as mathematics. Too many of you appear to have been hanging onto the mathematics, rather than, as was put earlier, (and I am sure we all remember being told), reading the question.
Well done PHILIP. The rest of you, detention I am afraid.
Can we all go to bed now?

"a triangle whose three sides AND its height are CONSECUTIVE whole numbers"
seems to me that the trap is to go for the obvious 3x4x5 Triangle and overlook the awkward height part of the clue?

I'm off to retire too.

Giddy was correct.

It is easily understood if you consider two right-angled triangles. The first has a base of 5, height 12 and hypotenuse 13 (25 + 144 = 169, area 5 x 12 / 2 = 30) and the second a base of 9, height of 12 and hypotenuse of 15 (81 + 144 = 225 and area 9 x 12 / 2 = 54). Put them alongside each other, joined on the side of length 12, and you have a single triangle of base 5 + 9 = 14 and sides of 13 and 15 with a height of 12 and an area of 30 + 54 = 84.

It just goes to show how ignorant (or is that arrogant) some maths teachers are. BTW I have an Honours degree in Mathematics. "Those that can, do; those that can't, teach."

Big Dave
Can I come out of the naughty corner now please?
TJ

TJ
Does the corner you're in form an equilateral triangle?

@ The Joker

Of course you can!

What's wrong with the following logic? A priori it's 2:1 on drawing a white ball. But once there are only two balls left in the bag and they are known to be 1 black 1 white the odds on drawing either in a further draw are 50:50.

Please ignore the above - just read the question again.

Although the question should read 'you now remove THE second ball'